Important Questions: Introduction to Three Dimensional Geometry | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.
Ans: Assume that the point Q (5, 4, –6) divides the line segment joining points P (3, 2, –4) and R (9, 8, –10) in the ratio k:1.
Therefore, by using the section formula, we can write it as:
(5, 4, -6) = [ (k(9) + 3)/(k + 1), (k(8) + 2)/(k + 1), (k(-10) - 4)/(k + 1)]
⇒ (9k + 3)/(k + 1) = 5
Now, bring the L.H.S denominator to the R.H.S and multiply it
⇒ 9k + 3 = 5k + 5
Now, simplify the equation to find the value of k.
⇒ 4k = 2
⇒ k = 2/4
⇒ k = ½
Therefore, the value of k is ½.
Hence, the point Q divides PR in the ratio of 1:2

Q2: Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.
Ans: Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is the z-coordinate of P, i.e., 8 units.

Q3: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Ans: Assume that P (x, y, z) be the point that is equidistant from two points A(1, 2, 3) and B(3, 2, –1).
Thus, we can say that, PA = PB
Take square on both the sides, we get
PA² = PB²
It means that,
(x-1)² + (y-2)² + (z-3)2 = (x-3)² + (y-2)² + (z+1)²
⇒ x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x2 – 6x + 9 + y² – 4y + 4 + z² + 2z + 1
Now, simplify the above equation, we get:
⇒ –2x –4y – 6z + 14 = –6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Hence, the required equation for the set of points is x – 2z = 0.

Q4: Prove that the points: (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right-angled triangle.
Ans: Let the given points be A = (0, 7, 10), B = (–1, 6, 6), and C = (–4, 9, 6)
Now, find the distance between the points
Finding for AB:
AB = √ [(-1-0)² + (6-7)² +(6-10)²]
AB = √ [(-1)² + (-1)² +(-4)²]
AB = √(1+1+16)
AB = √18
AB = 3√2 …. (1)
Finding for BC:
BC= √ [(-4+1)² + (9-6)² +(6-6)²]
BC = √ [(-3)² + (3)² +(-0)²]
BC = √(9+9)
BC = √18
BC = 3√2 …..(2)
Finding for CA:
CA= √ [(0+4)² + (7-9)² +(10-6)²]
CA = √ [(4)² + (-2)² +(4)²]
CA = √(16 + 4 + 16)
CA = √36
CA = 6 …..(3)
Now, by Pythagoras theorem,
AC² = AB² + BC² …..(4)
Now, substitute (1),(2), and (3) in (4), we get:
6² = ( 3√2)² + ( 3√2)²
36 = 18+18
36 = 36
The given points obey the condition of Pythagoras Theorem.
Hence, the given points are the vertices of a right-angled triangle.

Q5: If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then find the length of edges of a parallelopiped and the length of the diagonal
Ans: Let A = (2, 3, 5), B = (5, 9, 7)
To find the length of the edges of a parallelopiped = 5 – 2, 9 – 3, 7 – 5
It means that 3, 6, 2.
Now, to find the length of a diagonal = √(3² + 6² + 2²)
= √(9+36+4)
= √49
= 7
Therefore, the length of a diagonal of a parallel pole is 7 units.